Areciprocating single-acting pump has a cylinder of 125mm bore and 300mm stroke and draws water from a sump whose water level is 1m below the axis of the pump. the suction pipe is 2 m long and has a diameter of 50mm. calculate the speed in rpm at which separation occurs, if this takes place when the absolute pressure in the cylinder falls to 2.4m of water. atmospheric pressure is equivalent to 10.3m of water. assume simple harmonic motion of the piston. what is the maximum theoretical discharge in litres/min of this pump?

speed is 57.38 rpm

theoretical discharge =3.54 litres /min

Explanation:

given data

bore r = 125 mm

stock = 300 mm

water level = 1m

pipe L = 2 m

diameter = 50 mm

absolute pressure = 2.4 m

atmospheric pressure = 10.3 m

to find out

theoretical discharge and speed

solution

we will apply here pressure formula here

pressure (s) =  L/g × ( Area (p) / Area(s) ) × ω²×r×cos∅

for maximum pressure ∅ = 0

so put here all value

pressure (s) =  2/9.8 × ( π/4×125² / π/4×50² ) × ω²×300/2

pressure (s) = 0.191 ω²

and we know

atmospheric pressure = absolute pressure + pressure (s) + water level

10.3 = 2.4 + 1 + 0.191 ω²

solve it and we get

ω = 6 rad/sec

so discharge = A× L× N /60

so here N = 57.38

speed is 57.38 rpm

and

theoretical discharge = A(p) L N / 60

theoretical discharge = π/4×125² × 2 × 57.38 / 60

theoretical discharge =3.54 litres /min

answer:

50 vo; ts

explanation: